Let $R$ be an idempotent semiring and $ax=y$ and $by=x~forall ~x,yin R$, then show that $x=y$
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Here, a semiring $(R, +, .)$ is an idempotent in the sense that $x+x=x$ and $x.x=x~forall~xin R$. Let $R$ be an idempotent semiring and $ax=y$ and $by=x~text{if }xleq y~text{and }yleq x,~text{respectively }forall ~x,yin R$ and some $a, b$ in $R$, then show that $x=y$.
Note: Here, i am actually trying to show that the relation $leq$ on $R$ is an anti-symmetric relation with respect to the multiplicative operation on $R$. Further i have seen that if $R$ is a multiplicatively cancellative then $x=y$ can be easily verified but i don't need the $R$ to be cancellative.
semiring
asked Nov 28 '18 at 18:23
getegete
747
$endgroup$
add a comment |
$begingroup$
Here, a semiring $(R, +, .)$ is an idempotent in the sense that $x+x=x$ and $x.x=x~forall~xin R$. Let $R$ be an idempotent semiring and $ax=y$ and $by=x~text{if }xleq y~text{and }yleq x,~text{respectively }forall ~x,yin R$ and some $a, b$ in $R$, then show that $x=y$.
Note: Here, i am actually trying to show that the relation $leq$ on $R$ is an anti-symmetric relation with respect to the multiplicative operation on $R$. Further i have seen that if $R$ is a multiplicatively cancellative then $x=y$ can be easily verified but i don't need the $R$ to be cancellative.
semiring
asked Nov 28 '18 at 18:23
getegete
747
$endgroup$
$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32
add a comment |
$begingroup$
Here, a semiring $(R, +, .)$ is an idempotent in the sense that $x+x=x$ and $x.x=x~forall~xin R$. Let $R$ be an idempotent semiring and $ax=y$ and $by=x~text{if }xleq y~text{and }yleq x,~text{respectively }forall ~x,yin R$ and some $a, b$ in $R$, then show that $x=y$.
Note: Here, i am actually trying to show that the relation $leq$ on $R$ is an anti-symmetric relation with respect to the multiplicative operation on $R$. Further i have seen that if $R$ is a multiplicatively cancellative then $x=y$ can be easily verified but i don't need the $R$ to be cancellative.
semiring
asked Nov 28 '18 at 18:23
getegete
747
$endgroup$
Here, a semiring $(R, +, .)$ is an idempotent in the sense that $x+x=x$ and $x.x=x~forall~xin R$. Let $R$ be an idempotent semiring and $ax=y$ and $by=x~text{if }xleq y~text{and }yleq x,~text{respectively }forall ~x,yin R$ and some $a, b$ in $R$, then show that $x=y$.
Note: Here, i am actually trying to show that the relation $leq$ on $R$ is an anti-symmetric relation with respect to the multiplicative operation on $R$. Further i have seen that if $R$ is a multiplicatively cancellative then $x=y$ can be easily verified but i don't need the $R$ to be cancellative.
semiring
semiring
asked Nov 28 '18 at 18:23
getegete
747
asked Nov 28 '18 at 18:23
getegete
747
edited Nov 30 '18 at 15:37
gete
asked Nov 28 '18 at 18:23
getegete
747
asked Nov 28 '18 at 18:23
getegete
747
asked Nov 28 '18 at 18:23
getegete
747
747
$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32
add a comment |
$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32
$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32
add a comment |
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$begingroup$
So you define $x le y$ if there exists $a$ such that $ax = y$? And what does it mean when you write $ax=y$ and $by=x~forall ~x,yin R$?
$endgroup$
– Paul Frost
Nov 30 '18 at 18:19
$begingroup$
@Paul Frost $leq $ is a relation defined on $R$ such that $ forall~x, yin R,$ $ax=y $ whever $xleq y$ for some $ain R$. Now, to prove that the relation $leq$ is anti-symmetric, we have to assume that $xleq y$ and $yleq x$ and we need to show that $x=y$ using the relation as defined.
$endgroup$
– gete
Nov 30 '18 at 18:32